29. A carpenter is constructing a triangular roof for a storage shed as shown in the figure:
[The triangle has a 45-foot base and two 15-degree base angles -- dw]
Part A
How high will the peak of the roof rise above the top of the shed?
Give your answer to the nearest foot.
Part B
After the roof is constructed, it will be covered with an asphalt roofing material. The carpenter needs to calculate the combined length of the two sloping sides. What will be the total length needed of the roof covering?
Give your answer to the nearest foot.
Once again, I lament that all the good stuff is showing up on the test too late. By "the good stuff," I mean the material I covered in the Algebra II and Honors Integrated Math I classes I covered earlier this month (trig and constructions), and by "too late," I mean that I reach the problems after the week I was in that classroom. Just think about how many trig and construction problems I've covered after Friday the 13th, when I left that class!
For Part A, notice that we can't use any of the three trig functions until we have a right triangle. We form right triangles by drawing in the altitude to the base, which as we know, divides the isosceles triangle (which it is, as it has two base angles) into two congruent right triangles. The base of each triangle -- the leg adjacent to the 15-degree angle -- is 22.5 feet, and the value we are trying to find is the leg opposite the given acute angle. This implies that we need the tangent ratio:
a/22.5 = tan 15
a = 22.5 tan 15
a = 6.03, which is about 6 feet.
For Part B, we need to find the sum of the two hypotenuses of the right triangles. This implies that we need the cosine ratio:
22.5/c = cos 15
c = 22.5/cos 15
c = 23.3
2c = 46.6, which is about 47 feet.
As usual, we must deal with rounding error. It's so easy to round 23.3 down to 23 feet first and then double it to 46 feet, when we see that 46.6 rounding up to 47 feet is more accurate. As usual, PARCC counts both 46 and 47 feet as correct answers. On the other hand, students really have no business rounding 6.03 to anything other than six feet. And we can already see what the common student error will be -- giving 23 feet as an answer to Part B instead of doubling it to 46 or 47 feet. Another error is to plug 45 feet directly into the formulas instead of halving it to 22.5 feet.
Also, my rule of thumb is to use exact values from the question to solve Part B -- not the rounded off value from Part A. So we use the cosine in 22.5/c = cos 15, not the sine in 6/c = sin 15 or even the Pythagorean Theorem in 6^2 + 22.5^2 = c^2. In this problem, it probably wouldn't make that much of a difference since 6 and 6.03 are so close -- about half a percent difference.
Then again, using 6/c = sin 15 gives c = 23.2 (instead of 23.3) and so 2c = 46.4, which rounds to 46 instead of 47 feet. (Using 6 in the Pythagorean Theorem still gives 46.6 feet.) This doesn't matter for PARCC, which counts both 46 and 47 feet as correct, but it could make a difference in a real class where 46 feet could be marked wrong.
And of course, under no circumstances should we round 22.5 off in these formulas. If we tried to round 22.5 to 23, we obtain c = 23/cos 15 = 23.8 and 2c = 47.6, which rounds off to 48 feet -- and this answer will be marked incorrect.
By the way, notice that the total length of the roof covering is about 47 feet -- just two feet longer than the base of the roof -- yet it sticks up six feet in the air. That is, adding two feet to the roof makes it stick up six feet in the air. This unexpected result occurs because of the Pythagorean Theorem -- the equation a^2 + b^2 = c^2 implies that c = sqrt(a^2 + b^2) -- and when a is much smaller than b, c works out to be very close to b.
A classic question exploits this property of square roots -- suppose there is a track of length one mile, but when it gets hot, the track expands one inch, causing it to buckle. How high in the air will the track buckle? This is equivalent to the original question where the "base" of the roof (cold track) is one mile and the length of the roof "covering" (hot track) is one mile and one inch, and we wish to find the peak of the roof "covering" (hot track). The answer works out to be 15 feet -- and we'd definitely notice a track rising 15 feet in the air!
PARCC Practice EOY Question 29
U of Chicago Correspondence: Lesson 14-4, The Sine and Cosine Ratios
Key Postulates: Definition of Sine and Cosine
Common Core Standard:
Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems.
Commentary: Once again, this sort of question -- where we're given an isosceles triangle and we must cut it into two right triangles before we can apply trig -- is not in the U of Chicago text. I have run out of Pizzazz worksheets from the Algebra II class to apply, but I found an old worksheet from last year that covers another topic covered poorly in the U of Chicago text -- inverse trig functions. Both ideas -- applying trig to isosceles triangles and inverse trig -- are used to solve problems, so I decided to include last year's questions on today's worksheet.
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